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An important reaction in the formation of photochemical smog is the photodissociation of \(\mathrm{NO}_{2}\) : $$ \mathrm{NO}_{2}+h \nu \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g) $$ The maximum wavelength of light that can cause this reaction is \(420 \mathrm{~nm} .\) (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in \(\mathrm{kJ} / \mathrm{mol}\), that can be broken by absorption of a photon of \(420-\mathrm{nm}\) light? (c) Write out the photodissociation reaction showing Lewis-dot structures.

Short Answer

Expert verified
Light with a maximum wavelength of 420 nm falls within the visible light range, towards the violet end of the spectrum. The maximum bond strength that can be broken by this light is 285.33 kJ/mol. The Lewis-dot structures for the photodissociation reaction of nitrogen dioxide (NOâ‚‚) are as follows: \[ \begin{array}{ c c c c c c c } & \overset{\cdot}{N} = O - O \bullet & + & h\nu & \longrightarrow & \overset{\cdot}{N} = O & + & \overset{\cdot}{O} \\ & \text{NO}_{2} & & & & \text{NO} & & \text{O} \\ \end{array} \]

Step by step solution

01

Identify the part of the electromagnetic spectrum

Find where 420 nm falls within the electromagnetic spectrum. The range for visible light is about 380 nm (violet) to 750 nm (red). Therefore, 420 nm corresponds to the visible light range, toward the violet end of the spectrum.
02

Calculate maximum bond strength

Use the wavelength (λ) to find the energy of the 420-nm light, with Planck's constant (h) being approximately \( 6.626 \times 10^{-34} \) Js and the speed of light (c) being \( 3.00 \times 10^8 \) m/s: \[ E = \dfrac{hc}{\lambda} \] Convert the wavelength to meters: \( \lambda = 420 \times 10^{-9} \) m Now calculate the energy (E): \[ E = \dfrac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^8 \text{ m/s})}{420 \times 10^{-9} \text{ m}} \approx 4.742 \times 10^{-19} \text{ J} \] Finally, convert the energy to kJ/mol by using Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \): \[ \text{Energy in kJ/mol} = E * N_A * 10^3 = (4.742 \times 10^{-19} \text{ J})(6.022 \times 10^{23} \text{ mol}^{-1}) \times 10^3 = 285.33 \text{ kJ/mol} \] The maximum bond strength that can be broken by the 420-nm light is 285.33 kJ/mol.
03

Draw the Lewis-dot structures for the photodissociation reaction

Now we draw the Lewis-dot structures for the given reaction involving nitrogen dioxide (NOâ‚‚), nitric oxide (NO), and oxygen atom (O): \[ \begin{array}{ c c c c c c c } & \overset{\cdot}{N} = O - O \bullet & + & h\nu & \longrightarrow & \overset{\cdot}{N} = O & + & \overset{\cdot}{O} \\ & \text{NO}_{2} & & & & \text{NO} & & \text{O} \\ \end{array} \] The photodissociation of NOâ‚‚ into NO and an oxygen atom (O) with Lewis-dot structures is shown above.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photodissociation
Photodissociation is a critical process in the formation of photochemical smog. It involves the breakdown of molecules into smaller particles such as atoms or radicals through the absorption of light energy. In the given example, nitrogen dioxide (\(\mathrm{NO}_2\)) undergoes photodissociation when it absorbs a photon with sufficient energy. This results in the formation of nitric oxide (\(\mathrm{NO}\)) and an oxygen atom (\(\mathrm{O}\)).
The light required for photodissociation to occur must possess enough energy to break the molecular bonds. When light hits the molecule, the photon's energy breaks the chemical bond, creating new products. In the context of photochemical smog, photodissociation of nitrogen dioxide leads to the formation of reactive oxygen species, contributing to air pollution.
Electromagnetic Spectrum
The electromagnetic spectrum is a range of all types of electromagnetic radiation, which includes visible light, radio waves, gamma rays, and more. Light with a wavelength of \(420\ \mathrm{nm}\) falls within the visible light portion of the spectrum, specifically towards the violet end. Visible light ranges from about \(380\ \mathrm{nm}\) (violet) to \(750\ \mathrm{nm}\) (red).
  • Wavelengths towards the violet end of the spectrum possess higher energy compared to those closer to the red end.
  • In the context of the given reaction, the \(420\ \mathrm{nm}\) wavelength of light provides just enough energy to break certain chemical bonds in \(\mathrm{NO}_2\)
Understanding where light falls on the electromagnetic spectrum helps predict its energy and what kind of chemical reactions it might drive.
Bond Energy Calculation
Calculating bond energy is essential to understand the energy required to break a chemical bond. The bond energy calculation involves using the wavelength of light to determine the energy associated with a photon. Using the formula \(E = \frac{hc}{\lambda}\) where \(h\) is Planck's constant (\(6.626 \times 10^{-34}\ \mathrm{Js}\)) and \(c\) is the speed of light (\(3.00 \times 10^8\ \mathrm{m/s}\)), we can find the energy at given wavelengths.
For \(420\ \mathrm{nm}\) light, the energy is calculated as follows:
Convert \(\lambda\) to meters: \(420 \times 10^{-9} \ \mathrm{m}\)
Energy \(E = \frac{(6.626 \times 10^{-34}\ \mathrm{Js})(3.00 \times 10^8\ \mathrm{m/s})}{420 \times 10^{-9}\ \mathrm{m}} \approx 4.742 \times 10^{-19}\ \mathrm{J}\)
Converting to \(\mathrm{kJ/mol}\)
Using Avogadro's number \(6.022 \times 10^{23} \ \mathrm{mol}^{-1}\), the energy becomes \(285.33 \ \mathrm{kJ/mol}\).
This value indicates the maximum bond strength that a photon of \(420\ \mathrm{nm}\) light can break.
Lewis Structures
Lewis structures are diagrams that represent the valence electrons of atoms within a molecule. They provide insight into the molecular structure and the bonds between atoms.
In the photodissociation of nitrogen dioxide (\(\mathrm{NO}_2\)), Lewis-dot structures illustrate the breaking and formation of bonds.
  • The initial Lewis structure of \(\mathrm{NO}_2\) depicts a double bond between nitrogen and one oxygen atom, and a single bond with the second oxygen, with unpaired electrons.
  • After photodissociation, nitric oxide (\(\mathrm{NO}\)) forms, along with a single oxygen atom (\(\mathrm{O}\)) each retaining unpaired electrons.
These structures help visualize how molecules undergo changes during chemical reactions, such as photodissociation, displaying the distribution of electrons and the subsequent formation of new compounds.

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Most popular questions from this chapter

Nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) is the only important gaseous species in the lower atmosphere that absorbs visible light. (a) Write the Lewis structure(s) for \(\mathrm{NO}_{2}\). (b) How does this structure account for the fact that \(\mathrm{NO}_{2}\) dimerizes to form \(\mathrm{N}_{2} \mathrm{O}_{4}\) ? Based on what you can find about this dimerization reaction in the text, would you expect to find the \(\mathrm{NO}_{2}\) that forms in an urban environment to be in the form of dimer? Explain. (c) What would you expect as products, if any, for the reaction of \(\mathrm{NO}_{2}\) with CO? (d) Would you expect \(\mathrm{NO}_{2}\) generated in an urban environment to migrate to the stratosphere? Explain.

Why is the photodissociation of \(\mathrm{N}_{2}\) in the atmosphere relatively unimportant compared with the photodissociation of \(\mathrm{O}_{2}\) ?

The hydroxyl radical, \(\mathrm{OH}\), is formed at low altitudes via the reaction of excited oxygen atoms with water: $$ \mathrm{O}^{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{OH}(g) $$ (a) Write the Lewis structure for the hydroxyl radical (Hint: It has one unpaired electron.) Once produced, the hydroxyl radical is very reactive. Explain why each of the following series of reactions affects the pollution in the troposphere: (b) \(\mathrm{OH}+\mathrm{NO}_{2} \longrightarrow \mathrm{HNO}_{3}\) (c) \(\mathrm{OH}+\mathrm{CO}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{OOH}\) \(\mathrm{OOH}+\mathrm{NO} \longrightarrow \mathrm{OH}+\mathrm{NO}_{2}\) (d) \(\mathrm{OH}+\mathrm{CH}_{4} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3}\) \(\mathrm{CH}_{3}+\mathrm{O}_{2} \longrightarrow \mathrm{OOCH}_{3}\) \(\mathrm{OOCH}_{3}+\mathrm{NO} \longrightarrow \mathrm{OCH}_{3}+\mathrm{NO}_{2}\)

In 1986 an electrical power plant in Taylorsville, Georgia, burned \(8,376,726\) tons of coal, a national record at that time. (a) Assuming that the coal was \(83 \%\) carbon and \(25 \%\) sulfur and that combustion was complete, calculate the number of tons of carbon dioxide and sulfur dioxide produced by the plant during the year. (b) If \(55 \%\) of the \(\mathrm{SO}_{2}\) could be removed by reaction with powdered \(\mathrm{CaO}\) to form \(\mathrm{CaSO}_{3}\), how many tons of \(\mathrm{CaSO}_{3}\) would be produced?

(a) What is the primary basis for the division of the atmosphere into different regions? (b) Name the regions of the atmosphere, indicating the altitude interval for each one.

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